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Relativistic

Dipankar Chakrabarti Department of , Indian Institute of Technology Kanpur, Kanpur 208016, India (Dated: August 6, 2020)

1 I. INTRODUCTION

Till now we have dealt with non-relativistic . A free satisfying ~p2 Schrodinger has the non-relatistic E = 2m . Non-relativistc QM is applicable for with velocity much smaller than the velocity of (v << c). But for relativis- tic particles, i.e. particles with velocity comparable to the velocity of light(e.g., in atomic orbits), we need to use relativistic QM. For relativistic QM, we need to formulate a equation which is consistent with relativistic transformations(Lorentz transformations) of special . A characteristic feature of relativistic wave is that the of the particle is built into the theory from the beginning and cannot be added afterwards. (Schrodinger equation does not have any spin , we need to separately add spin wave .) it makes a particular relativistic equation applicable to a particular kind of particle (with a specific spin) i.e, a relativistic equation which describes scalar particle(spin=0) cannot be applied for a (spin=1/2) or vector particle(spin=1). Before discussion relativistic QM, let us briefly summarise some features of special theory of relativity here. Specification of an instant of t and a point ~r = (x, y, z) of ordinary defines a point in the space-time. We’ll use the notation

xµ = (x0, x1, x2, x3) ⇒ xµ = (x0, xi), x0 = ct, µ = 0, 1, 2, 3 and i = 1, 2, 3

x ≡ xµ is called a 4-vector, whereas ~r ≡ xi is a 3-vector(for 4-vector we don’t put the vector sign(→) on top of x. Consider two events in space-time (x0, x1, x2, x3) and (x0 + dx0, x1 + dx1, x2 + dx2, x3 + dx3) where x0 = ct so dx0 = cdt as c =velocity of light and is a constant. In three dimensional space we define the distance between two points. We generalize the notion of the distance between two points in space to the interval between two points in the space-time, say, ds. For ds to be same for all observer(ie, in all inertial frames), it must be invariant under Lorentz transformations and rotations. The interval is defined as

2 µ ν ds = gµνdx dx (1)

2 where gµν is the metric of the space-time. In   1 0 0 0       0 −1 0 0  g =   . (2) µν   0 0 −1 0      0 0 0 −1

So,

ds2 = (cdt)2 − ((dx1)2 + (dx2)2 + (dx3)2) = (cdt)2 − (dr~ )2 (3)

µ 0µ µ ν µ Under x transforms as x = Λ νx where Λ ν is a 4 × 4 representing the Lorentz transformation . For example, the operator for boost along x1 axis   γ −γβ 0 0       µ −γβ γ 0 0  Λ =   (4) ν    0 0 1 0     0 0 0 1

q where β = v/c and γ = 1/ 1 − (v/c)2. So, the transformed coordinates under the boost along x1: v v ct0 = γ(ct − x1), x01 = γ(x1 − ct), x02 = x2, x03 = x3. (5) c c Check that ds02 = (cdt0)2 − ((dx01)2 + (dx02)2 + (dx03)2) = γ2(cdt − βdx01)2 − γ2(dx1 − βcdt)2 − (dx2)2 − (dx3)2 = ds2 (6) i.e., ds2 is Lorentz invariant. ds2 can be both positive or negative unlike spatial distance (dr~ )2 which is always positive. If ds2 > 0 i.e., (cdt)2 > (dr~ )2, the interval is called "time-like" ds2 < 0 i.e., (cdt)2 < (dr~ )2, the interval is called "space-like" ds2 = 0 i.e., (cdt)2 = (dr~ )2, the interval is called "light-like".

3 covariant & contravariant vectors: Any quantity which transforms like xµ under Lorentz ∂ transformation is called a contravariant vector while anything which transforms like ∂xµ is called covariant vector. General convention for contravariant vector is aµ (i.e.,µ is in the superscript) ∂ and for covariant vector aµ (i.e, µ is in the subscript) i.e, ∂xµ = ∂µ. The inner product of a covariant vector and a contravariant vector is a Lorentz invariant(i.e., scalar). The contra and covariant vectors are related by

X ν xµ = gµνx . (7) ν

Using the convention of summation over repeated indices we can write the above eqn as xµ = ν ν µ µν gµνx where ν in gµν is repeated again in x and hence is summed over. Similarly, x = g xν. µν In Minkowski space, gµν = g . So, we have

ν 0 1 2 3 0 0 x0 = g0νx = g00x + g01x + g02x + g03x = g00x = x (8)

ν 0 1 2 3 1 1 x1 = g1νx = g10x + g11x + g12x + g13x = g11x = −x . (9)

2 3 Similarly x2 = −x and x3 = −x . Inner product or scalar product of two 4-vectors is defined as

µ 0 1 2 3 0 0 1 1 2 2 3 3 A · B = A Bµ = (A B0 + A B1 + A B2 + A B3) = (A B − A B − A B − A B ) (10) 0 0 ~ ~ µ ν µν = A B − A · B = gµνA B = g AµBν. (11)

Differential operators: ∂ 1 ∂ ∂ ∂ ∂ ∂ = = ( , , , ) (12) µ ∂xµ c ∂t ∂x1 ∂x2 ∂x3 1 ∂ = (∂ , ∂ , ∂ , ∂ ) = ( , ∇~ ) (13) 0 1 2 3 c ∂t 1 ∂ ∂µ = gµν∂ = ( , −∇~ ) (14) ν c ∂t The Lorentz invariant second order differential operator or the d’Alembertian operator is 1 ∂2 ∂2 ∂2 ∂2 1 ∂2 = ∂µ∂ = ( , −( + + )) = ( , −∇2). (15)  µ c2 ∂t2 ∂x2 ∂y2 ∂z2 c2 ∂t2 2 2 We know the relativistic mr = γm and energy E = mrc = γmc . The energy- 4-vector is pµ = (E/c, ~p) where ~p = γm~v. So, E (γmc2)2 p2 = g pµp = pµp = ( )2 − (~p)2 = − (γm~v)2 = m2c2 (16) µν ν µ c c2

4 (in the natural unit ~ = c = 1, p2 = m2). So, the relativistic energy momentum relation is given by E2 = (~p)2c2 + m2c4. Another useful quantity is

µ p · x = p xµ = Et − ~p · ~x. (17)

For non- (v << c), we can write

q ~p2 E = ~p2c2 + m2c4 = mc2(1 + )1/2 (18) m2c2 ~p2 (~p)4 ~p2 = mc2(1 + − + ··· ) = mc2 + − · · · (19) 2m2c2 8m4c4 2m

~p2 Negelecting the higher oredr terms, the of a non-relativistic particle is 2m = E − mc2.

A. Klein-Gordon Equation

Schrodinger proposed a relavistic form of his non-relativistic equation (at the same time when he developed his non-relativistic(NR) equation). Klein and Gordon developed this equation at a later time and is knaown as Klein-Gordon(KG) equation. Schrodinger used the NR energy- p2 momentum dispersion relation E = 2m . Using the correspondence ∂ E → Eˆ = i , ~p → ~pˆ = −i ∇~ (20) ~∂t ~

p2 in Eφ(~r, t) = 2m φ(~r, t), we arrive at the Schrodinger equation for . Now extend the same for relativistic particle with energy-momentum relation E2 = ~p2c2 + m2c4. So we get the relativistic

E2φ(x) = (~p2c2 + m2c4)φ(x) (21) ∂2 ⇒ − 2 φ(x) = (− 2c2∇~ 2 + m2c4)φ(x) (22) ~ ∂t2 ~  1 ∂2  m2c2 ⇒ − ∇~ 2 φ(x) = − φ(x) (23) c2 ∂t2 ~2 m2c2 ⇒ ( + )φ(x) = 0. (24) ~2 µ This equation is known as Klein-Gordon equation. Note that  = ∂µ∂ is a Lorentz invariant quantity, so the KG equation is Lorentz invariant only if φ is Lorentz invariant or Lorentz

5 scalar. Thus KG equation describes the of a scalar particle. The solution of the KG eqn is

φ(x) = Ne−i(Et−~p·~x) (25) √ where N is the normalization constant and energy E = ± ~p2c2 + m2c4 i.e., energy can be both positive and negative. : Pre-multiply Eq.(23) by φ∗(x) to get  1 ∂2  m2c2 φ∗(x) − ∇~ 2 φ(x) = − φ∗(x)φ(x) (26) c2 ∂t2 ~2 Now take the of Eq.(23) and post-multiply with φ(x), which gives 1 ∂2 m2c2 ( φ∗)φ − (∇~ 2φ∗)φ = − φ∗(x)φ(x) (27) c2 ∂t2 ~2 Eq(26)-Eq(27) gives: 1 ∂2φ 1 ∂2φ∗ φ∗ − φ − (φ∗∇2φ − φ∇2φ∗) = 0 (28) c2 ∂t2 c2 ∂t2 1 ∂  i  ∂φ ∂φ∗     ⇒ ~ φ∗ − φ + ∇~ · ~ φ∗∇~ φ − (∇~ φ∗)φ = 0 (29) c ∂t 2mc ∂t ∂t 2im 1 ∂ ⇒ ρ + ∇~ · ~j = 0 (30) c ∂t µ ⇒ ∂µj = 0 (31)

This is the continuity equation for the KG eqn, where i  ∂φ ∂φ∗  j0 = ρ = ~ φ∗ − φ (32) 2mc ∂t ∂t   ~j = ~ φ∗∇~ φ − (∇~ φ∗)φ . (33) 2im Recall the continuity eqn for Schrodinger equation, ρ is the density and ~j is the . Continuity equation has the interpretation of conservation of probability. It tells that if the probability of finding a particle in some region decreases, the probability of finding it out side that region increases, i.e., there is a flow of probability current so that the total probability remains conserved. Since the KG eqn also satisfies the same continuity eqn, it is natural to interpret ρ as the probability density and ~j as the probability current. [Note: Density

6 transforms like the 0th component of a 4−vector (jµ) under Lorentz transformation. Since φ is a Lorentz invariant quantity, φ2 does not transform like a density, but ρ defined in Eq.(32) does.] The probability density corresponding to the plane wave solution reads ρ = 2|N|2E. There are two major problems with the KG equation. (1) The eqn has both positive and solutions. The negative energy solution poses a problem! For large |~p| we can have large negative energy, i.e., the become unbounded from below. So, we can extract any arbitrary large amount of energy from the system by pushing it into more and more negative energy states. One may say, we truncate √ the physical space to be the positive energy states only i.e, only E = + ~p2c2 + m2c4 are physical. But then (a) the eigenstates don’t form a complete states, (both +ve and -ve energy states are Fourier modes of φ); if we don’t have completeness relation, we cannot have P too ie., we cannot expand a state χ in the basis of φ ( i.e., χ = i ciφi is no longer valid) and (b) a perturbation may cause the system to jump to a negative energy states. Since -ve energy states are valid solutions of the KG equation, we can not stop that. So, just interpreting negative energy states as unphysical does not . (2) The second problem is associated with the probability density. As we have seen ρ = 2|N|2E, i.e, ρ is negative if E is negative. But to interpret ρ as the probability density, it must be positive definite. [Though in QM, KG equation looks awkward at this moment, but in QFT this is a valid equation for scalar (spin=0)particles. Feynman and Stückelberg interpreted the positive energy states as particles propagating forward in time and negative energy states are propagating back- ward in time and thus represent propagating forward in time. But we’ll not discuss those developments here.]

7 B. :

The probability density in KG eqn depends on energy and becomes negative for negative energy. The energy in the expression of ρ appears due to the time derivative in Eq.(32). Dirac realised that this is due to the fact that KG eqn involves second order time derivative. Notice that Schrodinger equation invoves first order time derivative, and ρ does not involve any time derivative.. So, if we want to write a relativistic wave equation with positive definite probability density, the equation should be first order in time derivative. To be consistent with the Lorentz transformations in special theory of relativity, the wave equation with first order time derivative must also be first order in space derivatives. So, Dirac wrote the Hamiltonian as

2 H = α1p1c + α2p2c + α3p3c + βmc . (34)

Writing the momentum in differential operator form in the space, we must have the wave equation

∂ψ(x)  ∂ ∂ ∂  i = − i c(α + α + α ) + +βmc2 ψ(x) ~ ∂t ~ 1 ∂x1 2 ∂x2 3 ∂x3 2 = (−i~c~α · ∇~ + βmc )ψ(x) (35)

Since the above Hamiltonian has to describe a free particle, αi and β cannot depend on space and time, since such terms would have the properties of space-time dependent and

give rise to . Also αi and β cannot have space or time derivatives, the derivatives should

appear only in and E , since the equation is to be linear in all these derivatives. Thus αi, β are some constants. For relativistic particle, it must satisfy the relativistic energy momentum relation E2 = ~p2c2 + m2c4 i.e., it must satisfy the KG equation.

8 Squaring both sides of Eq.(35), we get

∂  ∂ ∂ ∂  (i )2ψ = − i c(α + α + α ) + +βmc2 ~∂t ~ 1 ∂x1 2 ∂x2 3 ∂x3  ∂ ∂ ∂  − i c(α + α + α ) + +βmc2 ψ ~ 1 ∂x1 2 ∂x2 3 ∂x3   2 2 2  2 2 2 ∂ 2 ∂ 2 ∂ 2 2 4 = − ~ c α1 + α2 + α3 + β m c ∂x12 ∂x22 ∂x32  ∂ ∂ ∂ ∂ ∂ ∂  − 2c2 (α α + α α ) + (α α + α α ) + (α α + α α ) ~ 1 2 2 1 ∂x1 ∂x2 1 3 3 1 ∂x1 ∂x3 2 3 3 2 ∂x2 ∂x3  ∂ ∂ ∂  −imc3 (α β + βα ) + (α β + βα ) + (α β + βα ) ψ (36) ~ 1 1 ∂x1 2 2 ∂x2 3 3 ∂x3

To satisfy E2 = ~p2c2 + m2c4, the above equation must satisfy

 2 2 2  2 ∂ 2 2 2 ∂ ∂ ∂ 2 4 −~ ( ) ψ = −~ c + + ψ + m c ψ (37) ∂t ∂x12 ∂x22 ∂x32

Now if Eq.(36) has to satisfy Eq.(37), then αi (i = 1, 2, 3) and β must satisfy

αiαj + αjαi = 0, (i 6= j) (38)

αiβ + βαi = 0 (39)

2 2 αi = 1, β = 1 (40)

Clearly, αi and β cannot be ordinary classical numbers, rather they anticommute with each other. So, Dirac propopsed that they are matrices. The above anticommutation relations can be writen in the short forms as

{αi, αj} = 0 (i 6= j), {αi, β} = 0 (41)

2 (The notation { , } is called the anticommutator.) Combining with the fact that αi = 1 we can write

{αi, αj} = 2δijI. (42)

If αi and β are matrices, ψ cannot be a single component , it must have more than one components that can be written as a vector on which the matrices should operate. For Dirac equation, we need four linearly independent matrices satisfying the anticommu-

tation relations. Since the Hamiltonian is hermitian, each of the four matrices αi, β must be

9 hermitian and hence they are square matrices(n × n). Since squares of all four matrices are

unity, their eigenvalues are +1 and −1. If we choose β to be diagonal, then αi cannot be diagonal as they anticommute with β. In 2 , we have three which anticommute with each other but the fourth linearly independent matrix that we can have in 2D is the which commutes with all other matrices. So, we cannot find a linearly independent fourth matrix to anticommute with the Pauli matrices. Similarly, we fail to find four 3 × 3 matrices to satisfy all the above conditions. The smallest possible to have four such matrices is 4 × 4. One such of matrices are:     0 σ I 0  i   αi =   , β =   (43) σi 0 0 −I

where σi are the Pauli matrices and I is 2 × 2 identity matrix.       0 1 0 −i 1 0       σ1 =   , σ2 =   , σ3 =   , (44) 1 0 i 0 0 −1

αi and β are not unique. All matrices related to these matrices by any unitary 4 × 4 matrix are equally valid i.e.,

0 −1 0 −1 αi = UαiU , β = UβU (45)

will also satisfy the Dirac equation and all the anticommutation relations. Since αi and β are 4×4 matrices, ψ is a 4-component column vector. As UU −1 = I, you can show that for Lorentz invariance of the Dirac equation, ψ then transforms as ψ0 = Uψ. Free particle solution: Like KG equation, we look for the solution in which the space-time behaviour is of plane wave form:

−i p·x −i Et +i ~p·~x ψ(x) = ωe ~ = ωe ~ ~ . (46) where ω is a 4-component vector, indepndent of x and is called the Dirac . Let us write ω in 2-component notation   φ   ω =   (47) χ

10 where φ and χ are 2 -component . Putting the solution in the Dirac equation (Eq.35), we get       φ φ φ     2   E   = c~α · ~p   + βmc   χ χ χ         0 c~σ · ~p φ I 0 φ     2     =     + mc     c~σ · ~p 0 χ 0 −I χ     mc2I c~σ · ~p φ     =     . (48) c~σ · ~p − mc2I χ

The matrix equation can be written as two coupled equations:

Eφ = mc2φ + c~σ · ~pχ ⇒ (E − mc2)φ = c~σ · ~pχ, (49)

and

Eχ = −mc2χ + c~σ · ~pφ ⇒ (E + mc2)χ = c~σ · ~pφ c~σ · ~p ⇒ χ = φ. (50) E + mc2 Putting Eq.(50) in Eq.(49) we have c~σ · ~p (E − mc2)φ = c~σ · ~p φ E + mc2 c2(~σ · ~p)2 ~p2c2 = φ = φ (51) E + mc2 E + mc2 where we have used (~σ · A~)(~σ · B~ ) = A~ · BI~ + i~σ · (A~ × B~ ) ⇒ (~σ · ~p)2 = (~p)2. So finally we get,

(E − mc2)(E + mc2)φ = ~p2c2φ ⇒ E2 = ~p2c2 + m2c4 (52) √ i.e, E = ± ~p2c2 + m2c4, which means negative energy solutions are still admitted. Dirac’s prescription cannot get rid of the negative energy solutions. Let us postpone the discussion on negative energy now. We’ll come back to the issue of negative energy solution at the end. Let us first check what happens to the probability density. To derive the continuity equation, first premultiply the Dirac equation by ψ† : ∂ψ ψ†i = ψ†(−i c~α · ∇~ + βmc2)ψ (53) ~ ∂t ~

11 Take hermitian conjugate of the Dirac equation and post multiply with ψ :

∂ψ† ← −i ( )ψ = ψ†(i c~α· ∇ +βmc2)ψ (54) ~ ∂t ~

← † † † Note that the spatial derivative ∇ acts on the left i.e.,on the ψ and αi = αi and β = β. Now subtracting Eq.54 from Eq.53, we get

∂ ← i (ψ†ψ) = −i c ψ†(~α · ∇~ + ~α· ∇)ψ = −i c∇~ · (ψ†~αψ). (55) ~∂t ~ ~

Thus we get the continuity equation( in the covariant form)

1 ∂ (ψ†ψ) + ∇~ · (ψ†~αψ) = 0 (56) c ∂t 1 ∂ ⇒ ρ + ∇~ · ~j = 0 (57) c ∂t µ ⇒ ∂µj = 0 (58) where ρ = j0 = ψ†ψ and ~j = ψ†~αψ. Since ψ is a 4-component vector, let us write   ψ  1     ψ2 ψ =   . (59)   ψ   3   ψ4

Now the probability density   ψ  1     ψ2 † ∗ ∗ ∗ ∗   2 2 2 2 ρ = ψ ψ = ψ ψ ψ ψ   = |ψ1| + |ψ2| + |ψ3| + |ψ4| ≥ 0 (60) 1 2 3 4   ψ   3   ψ4

⇒ ρ is positive definite. Thus it can be interpreted as probability density. But now we need to interpret ψ which is a four component vector. What is the significance

or physical meaning of these components? Note that the α matrices involve Pauli matrices σi. ~ ~ We know that the spin operator are written as S = 2~σ. So, one obvious question arises: Do the different components in the represent different spin components?

12 let us consider the positive energy solution only. From Eqs. (47 and 50), we can write   φ ω =   . (61)  c~σ·~p  E+mc2 φ The 2-component spinor φ is completely arbitrary. We may choose two linearly independent forms     1 0     φ↑ =   , φ↓ =   (62) 0 1

~ These are the eigenstates of Sz = 2 σz. The most general form can be expanded in terms of these two basis vectors   a   φ = aφ↑ + bφ↓ =   . (63) b

We have two linearly independent solution for any energy E. So, for a given 4-momentum, there are just two linearly independent solutions ie, 2-fold degenerate solutions for ω, just as expected for a quantum system with j = 1/2 ( (2j + 1) = 2). To give the spin interpretation let us consider the rest frame of the particle, i.e., ~p = 0. Then E = mc2(we are considering positive energy only). Two linearly independent solutions in the rest frame can be written as     1 0             0 −imc2t/ 1 −imc2t/ ψ =   e ~, ψ =   e ~, (64) 1   2   0 0         0 0

These are of the operator   σz 0 Σ = ~   (65) z 2   0 σz

~ with eigenvalues ± 2 . Both solutions have same energy, but eigenvalues of Σz distinguishes 2 them. Similarly for negative energy solutions (E = −mc ), we can have two solutions ψ3

and ψ4 with 1 in the place third and fourth element in the column matrix respectively and

13 corresponding sign change in the exponential. Generalizing the definition of the operator Σz for all three components, we write   ~σ 0 ~ ~   Σ =   . (66) 2 0 ~σ

They satisfy the standard commutation relation for spin operators [Σx, Σy] = i~Σz. Thus Σ is 1 the appropriate for the spin 2 operator to our rest frame solution and we may conclude that atleast in the rest frame Dirac solution represents spin-1/2 particles. We know that spin is a fundamental property of a particle, spin of a particle does not change if we boost the system. So a spin-half particle in rest frame is a spin-half particle in all frames. Thus Diarc equation describes the dynamics of spin-half particles or . But for ~p 6= 0, Σ is no longer a suitable operator to describe spin, as it does not commute withH = c~α · ~p + βmc2. We need to find an operator that commutes with H and whose eigenvalues distinguish the two states with same energy. One of such operator is the helicity operator   ~σ·~p 0 h(p) = ~  |~p|  . (67) 2  ~σ·~p  0 |~p| Physically the helicity operator h(~p) gives the projection of spin(Σ) along the direction of ~p.

~ Eigen values of h(p) are called helicity of the particle. h(p) has the eigenvalues of h = ± 2 i.e.,

h(p)ω = ±~ω 2       ~σ·~p 0 φ φ ⇒ ~  |~p|    = ±~   . (68) 2  ~σ·~p   c~σ·~p  2  c~σ·~p  0 |~p| E+mc2 φ E+mc2 φ For positive helicity, we must have ~σ · ~p φ = φ , (69) |~p| + + and for negative helicity, ~σ · ~p φ = −φ , (70) |~p| − −

φ+ and φ− are linearly independent helicity spinors.

14 Lorentz transformaion properties of Dirac spinor: Rotation: For simplicity consider the particle is moving along z-direction i.e., ~p = (0, 0, p). Then we can simply choose     1 0     φ+ =   , φ− =   (71) 0 1 Now rotate the coordinate system about x axis by an angle θ. Since ~p is a vector, it transforms just the same way as ~r.

0 px = px = 0, (72)

0 py = cos θ py + sin θ pz = sin θ p (73)

0 pz = − sin θpy + cos θ pz = cos θ p (74)

0 Let under rotation φ+ → φ+, so we must have ~σ · p~0 φ0 = φ0 (75) |p~0| + + since it must represent a positive helicity state, the state is merely being described in a different coordinate system. Writing the above equation in explicit form, σ p0 + σ p0 + σ p0 x x y y z z φ0 = φ0 (76) |p0| + +     1 0 −i sin θ cos θ 0      0 0 or, p   + p   φ+ = φ+ (77) p i sin θ 0 0 − cos θ   cos θ −i sin θ   0 0 or,   φ+ = φ+. (78) i sin θ − cos θ   a 0   let φ+ =  , so we can write the above equation as b a cos θ − ib sin θ = a (79) ia sin θ − b cos θ = b. (80)

Solving the above two coupled equations, we get a = cos(θ/2) and b = i sin(θ/2), i.e.,     cos(θ/2) cos(θ/2) i sin(θ/2) 0     φ+ =   =   φ+. (81) i sin(θ/2) i sin(θ/2) cos(θ/2)

15 Similarly we can write the solution for negative helicity state as     i sin(θ/2) cos(θ/2) i sin(θ/2) 0     φ− =   =   φ−. (82) cos(θ/2) i sin(θ/2) cos(θ/2)

Thus, the wavefunctions in the rotated coordinate system are linear combinations. of the corresponding components in the original coordinate system. The transformation is given by a

2 × 2 matrix as the spinors(φ±) are 2-component vectors. The matrix elements of the rotation operator depends on the rotation angle but in the spinor case it is the half-angle θ/2 that enters. (Note: In case of vector, the rotation angle θ appears in the ). We can write the above rotation in a compact form as

φ0 = eiσxθ/2φ. (83)

For rotation about an arbitrary direction nˆ, the transformation rule can be generalized to

φ0 = ei~σ·nθ/ˆ 2φ = Uφ. (84)

So, spinors have a well-defined transformation rule under rotation. But it is different from vector, or scalar. Since the rotation operator U is unitary , the of the state is preserved: φ0†φ0 = φ†φ. Boost: Let in the S-frame the system is at rest i.e., ~p = 0 and energy E = mc2 and in S0 ~0 0 0 frame, it has momentum p = (px, 0, 0) and energy E . That is, we are considering a Lorentz 0 0 0 transformation where S frame has velocity (−vx, 0, 0) with respect to S frame where vx = px/E . pµ = (E/c, ~p) transforms like a 4-vector. For a boost along x-axis

E0 E = cosh ξ + sinh ξp (85) c c x E p0 = sinh ξ + cosh ξp (86) x c x where tan ξ = p0 /E0. x   1         1 0 In the rest frame (S-frame), we choose φ =   ie., ω =  .     0 0     0

16 In S0 frame the spinor is     1 1             0  0   0  ω = N    = N    (87)      0 1   0 0   cσxpx    cpx    E0+mc2    E0+mc2   0 1 where N is the normalization constant. Lorentz transformation can be considered as a rotation by an imaginary angle ξ = iθ. But φ and χ transform differently under the boost transformation due to the fact that E and ~p enter differently in φ and χ. So, we expect , compared to rotation, a pure imaginary angle and a matrix that couples φ and χ. It can be written as

ω0 = eαxξ/2ω (88)   0 σ  x where αx =  , so σx 0

  cosh ξ/2 σx sinh ξ/2 αxξ/2   e = cosh ξ/2 + αx sinh ξ/2 =   (89) σx sinh ξ/2 cosh ξ/2

2 as αx = 1. For a boost in any arbitrary direction the general transformation rule is:  ~α · ~v ξ  ω0 = exp ω. (90) |v| 2 Thus the Lorentz transformation rules of ω or ψ is different from a vector. So, ψ is not a vector under Lorentz transformation, it is called a spinor. Any object which transforms like ψ under Lorentz transformation is called a spinor. √ Positive and negative energy spinors: Write E = + ~p2c2 + m2c4. As time derivative in Dirac equation gives E = p0c, we can write the Dirac equation as       φ mc2I c~σ · ~p φ 0       p c   =     . (91) χ c~σ · ~p −mc2I χ

For positive energy solution E = p0c > 0 and the Dirac spinor is written as   φs ωs = N   (92)  c~σ·~p s E+mc2 φ

17 where s = 1, 2 corresponds to different spin states. Choosing the normalization condition ω†ω = 2E, we can write the positive energy spinor   √ φs u(p, s) = E + mc2   , (93)  c~σ·~p s E+mc2 φ and the complete plane wave positive energy solution is

ψs = u(p, s)exp−ip.x/~ (94) where pµ = (+E/c, ~p). √ For negative energy states, p0c = − ~p2c2 + m2c4 = −E < 0. To construct the spinor for -ve energy, let us first consider the solution at rest frame: p0c = −E = −mc2, ~p = 0(we assume E is a positive quantity). Then the Dirac equation simplifies to       φ mc2I 0 φ 2       −mc   =     (95) χ 0 −mc2I χ

  0   which gives φ = 0, ⇒ ω =  . For finite momentum, we have from the Dirac eqn χ

      φ mc2I c~σ · ~p φ       −E   =     . (96) χ c~σ · ~p − mc2I χ

Which gives,

c~σ · ~p φ = − χ (97) E + mc2

  − c~σ·~p χs  E+mc2  ⇒ ω(−E, ~p,s) =   (98) χs

µ Let us now change the sign of ~p, i.e, consider the solution for negative 4-momentum p− = (−E/c, −~p) = −pµ,   c~σ·~p χs  E+mc2  ω(−E, −~p,s) = N   (99) χs

18 Again using the same normalization condition ω†ω = 2E, we write the negative energy spinor as   √ c~σ·~p χs 2  E+mc2  v(p, s) = E + mc   (100) χs s = 1, 2 and the complete plane wave solution reads

ψs = v(p, s)e−ip−·x/~ = v(p, s)eip·x/~ (101) where pµ = (E/c, ~p). Dirac’s interpretation of the negative energy solutions: The physical interpretation of positive energy solutions is straight forward. They describe 1 m spin- 2 particles with 4-momentum p u = (E/c, ~p). The probability density ρ and the probabil- ity current ~j both are positive definite. But since the negative energy solutions are also allowed, like KG equation, a particle with +ve energy can cascade down through the -ve energy levels without limit. Hence +ve energy states cannot be stable! To make any sense of Dirac equation, one then needs to make the +ve energy states stable, preventing them to make transition to -ve energy states. Here comes the masterclass of Dirac! Dirac postulated that the normal empty or state corresponds to the state with no positive energy particle and all the negative energy states are completely filled up! The state with completely filled up negative energy levels is called the . Since Dirac eqn describes fermions, according to Pauli exclusion principle only two electrons(one spin up and one spin down) can occupy an and once they are occupied any +ve energy particle is forbidden to fall in the -ve energy levels. Let us assume that the spin 1/2 particle we are talking about is an . So, the vacuum is the state where all negative energy levels are filled up by electrons i.e., has infinite negative and energy! But since all represent finite fluctuations of charge and energy with respect to the vacuum, it leads to an acceptable theory and we rescale the vacuum to be without any charge and energy (charge of the vacuum=0, energy of the vacuum=0, spin of the vacuum=0). Assume an electron with energy −E and spin up is removed from the Dirac sea. It will create a "" relative to the normal vacuum: energy of the "hole" = −(−E) = +E → positive charge of the "hole" = −(−e) = e → positive charge.

19 FIG. 1: Energy levels of Dirac equations.The blobs represent electrons occupying the energy levels in Dirac sea. Each level can have two electrons(spin up and spin down) according to Puali exclusion principle. The open circle represents the absence of an electron i.e., presence of a "hole". spin of the "hole"=-(up)= down. Thus the absence of a negative energy electron with spin up is equivalent to the presence of a positive energy and positively charged "hole" with spin down. So, "hole" represents the an- tiparticle of the electron(i.e, ). So, the unfilled negative energy states according to Dirac, represent positive energy antiparticles. Thus in order to give stability to the +ve energy states, Dirac predicted the existence of positron!(Actually, when Dirac wrote this equation, positron was not known and he thought which is a positively might be the of electron!). Anderson discovered positron in 1932 to win the Nobel prize. DIrac required to consider infinite number of electrons filling up the negative energy states to describe a stabe "single" electron with positive energy! So, in that sense, it is no-longer a "single-particle" theory! Exciting a negative energy electron to a positive energy state, ie. creating a physical electron from the vacuum also creates a "hole" in the Dirac sea or a positive energy positron, which corresponds to the process of creating an e−e+ pair! Appropriate theory to describe the particle creations or destruction is the Quantum Theory! Dirac’s theory thus suggests to move to quantum field theory.

20 Here we have discussed only free relativistic equations. If we consider Dirac equation in for ( i.e., Dirac equation in central potential), the fine structures are observed in the energy eigenvalues. One can also include interaction with radiations by introducing the gauge fields as we have discussed before for Schrödinger eqn. But we’ll not discuss them here.

[1] Gauge Theories in , Volume -1, From Relativistic Quantum Mechanics to QED., by Aitchison and Hey. [2] Quantum Mechanics, by L.I. Schiff

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